Is the electric field in matter affected by polarization?

[rayveldkamp]

Hi,
Suppose we have a polarized object, and wish to calculate D(r) using
D(r) = epsilon E + P

Do we have to account for the electric field that the polarized object creates, or is it enough to just use the electric field which created the polarization?
Thanks

Ray

 

[Meir Achuz]

That equation relates the total E field to the total D field, so you should include the field due to the object.

 

[Antiphon]

Hi,
Suppose we have a polarized object, and wish to calculate D(r) using
D(r) = epsilon E + P

Do we have to account for the electric field that the polarized object creates, or is it enough to just use the electric field which created the polarization?
Thanks

Ray

Your question is a touch confusing, and the equation is incorrect.

The correct equations which both apply to your question are:

$$D = \epsilon E$$ and
$$D = \epsilon_0 E + P$$

In the first equation, the E is the actual E in the material but the epsilon
is related to the material, not to free space. In the second equation, the
epsilon is the epsilon of free space but the E is not the E which you would
measure in the material.

The object will have a different electric field inside it than the electric
field which would have been present if the object were not there.
The D vector does not change whether the mateiral is there or not.

 

[Meir Achuz]

I assumed that Ray did leave the subscript 0 out of the epsilon.
The E is the same in each of Antiphon's trwo equatiopns.
The equations are just related by the connection between epsilon and chi, the electric susceptibillity.

 

[rayveldkamp]

Hi,
Thanks for the replys, it turned out i only needed to include the field that created the polarization. Help was appreciated.
Thanks

Ray

 

[Antiphon]

Meir is right. It's the same E in both equations.